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Chapter 31: Problem 73

Show that identical objects placed equal distances on either side of the focal point of a concave mirror or converging lens produce images of equal size. Are the images of the same type?

Short Answer

Expert verified
Using the thin lens (or mirror) equation, it can be derived that both objects would create images of the same size due to being identical and placed the same distance from the focal point. However, as one object is placed on the virtual side, its image appears to be upright (thus virtual), and the other object, being on the real side, forms an inverted real image. Therefore, both images are not of the same type.

Step by step solution

01

Set up the lens/mirror equation

Use the thin lens (or mirror) equation, \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Here, \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance.
02

Apply for two objects placed at identical distance on either side of the focal point from the mirror/lens

Both objects are identical (therefore same height, \(h_o\)) and placed the same distance \(d_o\) from the focal point. Therefore, applying the lens/mirror equation for both should lead to identical values for the image distance \(d_i\). Considering one to be positive and the other to be negative, we have objects located at +d_o and -d_o
03

Determine the sizes of the images

The magnification (magnifying power), \( m \), of a mirror/lens is the ratio of the height of the image, \( h_i \), to the height of the object, \( h_o \). The equation is: \( m = -\frac{d_i}{d_o} = \frac{h_i}{h_o} \). Rearrange this to find \( h_i = m * h_o = -\frac{d_i}{d_o} * h_o \). Since the magnification factors (and object heights) are equal, we have, \( h_{i1} = h_{i2} \)
04

Determine the nature of the images

Consider both cases: the object distance \( d_o \) is positive and negative. If \( d_i \) is positive, then the image is real and inverted; if \( d_i \) is negative, then the image is virtual and upright. In our case, since we have both a positive and negative \( d_o \) (with corresponding \( d_i \)), we'll have both a real and virtual image.

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Most popular questions from this chapter

Generalize the derivation of the lensmaker's formula (Equation 31.7 ) to show that a lens of refractive index \(n_{\text {lens }}\) in an external medium with index \(n_{\mathrm{ext}}\) has focal length given by $$\frac{1}{f}=\left(\frac{n_{\mathrm{lens}}}{n_{\mathrm{ext}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$

Two specks of dirt are trapped in a crystal ball, one at the center and the other halfway to the surface. If you peer into the ball on a line joining the two specks, the outer one appears to be only one-third of the way to the other. Find the refractive index of the ball.

At what two distances could you place an object from a \(45-\mathrm{cm}-\) focal-length concave mirror to get an image 1.5 times the object's size?

(a) Find the focal length of a concave mirror if an object placed \(38.4 \mathrm{cm}\) in front of the mirror has a real image \(55.7 \mathrm{cm}\) from the mirror. (b) Where and what type will the image be if the object is moved to a point \(16.0 \mathrm{cm}\) from the mirror?

Chromatic aberration results from variation of the refractive index with wavelength. Starting with the lensmaker's formula, find an expression for the fractional change \(d f / f\) in the focal length of a thin lens in terms of the change \(d n\) in refractive index.
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